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In each revoution, the tire travels  pi * diameter =  11.5 pi  ft      (1)

6 mlies  = 6 * 5280  ft   (2)

So....the number of revolutions that the tire makes is   (2)  / (1) =

[ 6 * 5280 ]   /[  11.5 * pi  ] ≈  877 revolutions

900\( \)

In triangle ABC  since angles BAC and BCA  = 35°

Then angle ABC  = 180  - 35  - 35   =  180 -70 =110°

And because ABC  is an inscribed angle....then the major arc it intercepts, ADC,  has twice its measure = 220°

 So  minor arc ABC  = 360 - 220  =  140°

And since  CDA is an inscribed angle  intercepting this arc, it has 1/2 of its measure = 

(1/2) 140   =  70°

  x + 4

__________

x^2 + ax + b

If the vertical asymptotes are   x  =1  and x = -2....the denominator  is

( x - 1)  ( x +  2)   =  x^2  + x  -  2

So.....

a =  1    and b  = -2

So....a +  b  =   1  +  (-2)  =   -1

Sorry...see my corrected answer  !!!

Ratio of Perimeters   =   7: 9   =  (scale factor)

Ratio  of Areas  =  (scale factor)^2 =  (7 : 9 )^2  =  49 : 81

Area of  Red Hexagon  = (Area of Large Hexagon)  * ( scale  factor)^2   =

                                           (210)  * ( 49/81)  ≈   127  in^2

\(\approx4.77\) percent

Unfortunately your answer was incorrect; although, I do have another chance to answer the question if you could perhaps further help

Ratio of side  lengths    Ratio  of Perimeters    Ratio of Areas

(Scale Factor)                (Scale Factor)            (Square of Scale Factor )

        7 : 12                           7 : 12                        49 : 144

       32 : 44                         32: 44                       1024 : 20736

The area  of the larger circle is  pi * radius^2  = pi * 6^2  = 36 pi  (1)

The radii of the two smaller   circles  = 3

So...the area of these two circles   is   2 * pi * (3)^2  =  18 pi    (2)

So...the shaded area is  (1) - (2)   =  [ 36 - 18 ] pi  =   18 pi  [ units^2 ]

I just used several values for n and all of them show that your expression is TRUE!!. But I don't know how to prove it! Sorry about that.

Draw  radius OC

And since  OC = OA, then angles OCA  and OAC  are equal 

Then angle  COA  =  180   - 18  -18 = 180  - 36  = 144°

And by symmetry, angle BOC = 144°

So angle AOB  = 360  - 144 - 144  = 360 - 288  = 72°

we are looking fro x interrcepts which are (2,0) and (5,0). so x will be 5 and 2.

Assuming that these are similar figures...and that the side of the larger triangle with a length of 6  corresponds to the side of the smaller  triangle with a length of 2

The  scale factor of the larger triangle to the smaller  is just  the above ratio  = 6/2  = 3/1

This is the linear scale factor for  any dimension....all corresponding sides of the larger figure to the smaller  wil have this ratio...also...the height of the larger triangle to the smaller wioll also have this ratio

Perimeter is a linear measure...so ...the ratio of the perimeters will just  be  the same  ratio    3 : 1  

The ratio of the areas will be  the square of the scale factor  = (3/1)^2  = 9/1  = 9 : 1

So...the area of the the blue triangle will be :

Area of the smaller  triangle * (scale factor)^2  =

2  * (3/1)^2  =

2 * 9  =

18 ft^2

To  see this last part a litle more clearly

Note that the area of the smaller triangle  = (1/2) base * height  = (1/2) b * h

But since each dimension of the larger triangle is 3 times that of the smaller...its area  is

(1/2) ( 3* base of smaller triangle) ( 3 * height of smaller triangle)  =

(1/2) (3 b) (3 h)  = (1/2) 9 bh  =  9 [ (1/2) bh ] =   9 times [ the area of the smaller triangle]

Does that make some sense ???

Ok let’s say w and z equals -1. Imput it in.

 1.         |-1| = |-1| = 1. It works since the absolute value of -1 is 1

 2.         -1*-1=1 because negatives mutiplied by negatives is a positive.

the third one I’m going to do it top then bottom.

 3.            top: -1 + -1 = -2

bottom: 1 + (-1 * -1) = 1 + 1 = 2

together: -2 /2 = -1 and -1 is a real number. (Unless you mean it has to be a positive number)

198/88=2.25 

so 10*2.25=22.5

The area of the larger circle  will be  4 pi

So...the  shaded area  is  =  (5/12) * 4 pi  =  20/12 pi  =  ( 5/3 ) pi 

Let  the measure  of  the smaller angle   ADC  = theta  (in rads)    

And let the measure of the larger angle ADC  = [2pi - theta ]  in rads

So...the  area of the sector subtended by the large circle  is (1/2) 4^2 * θ =  2θ

And let the area of the sector subtended by the larger circle be  (1/2) * 1^2 * (2pi - θ) =

(1/2) (2pi - θ)  

So we have that

2θ  +  (1/2)(2pi - θ)  =  (5/3)pi

2θ + pi - θ/2  =  (5/3)pi

(3/2)θ =  2/3 pi

θ = (2/3)(2/3) pi  =  (4/9) pi  

So...the smaller angle ADC  =  (4/9)pi * 180 / pi  =   20 * 4   = 80°

Proof

Area of  shaded sector of lager  circle  = 2(4/9) pi  =  8/9 pi

Area of shaded  sector of smaller circle = (1/2) (2pi - (4/9) pi)  = (1/2) (14/9) pi = (7/9) pi

Combined areas =  (8/9 + 7/9) pi  =  (15/9) pi  =  (5/3) pi

We have an equilateral  triangle...

The area of this triangle  is  3(1/2) * 8^2  * sin (120°) =  3 (32) * √3 / 2  =  3 * 16 * √3  =

48√3  cm^ 2     (1)

The area of the circle  = pi * 8^2  =  64 pi   cm^2     (2)

So...the shaded area  is  (2)  - (1)  =  [ 64 pi  - 48√3 ] cm^2   =   "A"

Well, thanks to heureka, we can use his wonderful formula for this instead of actually counting them!!.
[(n - 1) x (b - 1) x (b^n - 1)] / b, where n=number of digits, and b=base used.
[(5 - 1) x (8 - 1) x (8^(5 - 1))] / 8 =
[   4      x      7     x    4,096] / 8   =
                114,688 / 8                  =14,336 zeros between 10,000 and 77,777 in base 8.                  [Amazingly beautiful formula !!!. Thanks heureka]

I hope you get the C you wanted.

10pi =75 x pi/180 x r(radius), solve for r

10pi = 5/12 pi x r    multiply both sides by 12/5 pi

r =12/5 pi x 10 pi

r = 24 - Radius of this circle.

good luck

lets find the total circumference. 10\pi*4.8= 48\pi

we know the circumference is 2pi*r

so 48pi/2pi=24

r=24

16-4=12 AC=sqrt(12)=2sqrt(3)

ABC area is 2sqrt(3)

4\pi-2sqrt(3) is the area of the shaded region